Class XII-Science : Physics, BoardPaper 2013, Set-3 “All India”

Ge ne ral Instructio ns:Ge ne ral Instructio ns:

(i) All questions are compulsory.

(ii) There are 29 questions in total. Question Nos. 1 to 8 are very short answer type questions

and carry one mark each.

(iii) Questions Nos. 9 to 16 carry two marks each. Questions Nos. 17 to 25 carry three marks

each and questions Nos. 27 to 29 carry five marks each.

(iv) There is no overall choice. However, an internal choice has been provided in one

question of two marks, one question of three marks and all three questions of five marks

each. You have to attempt only one of the choices in such questions.

(v) Questions No. 26 is value based question carries four marks.

(vi) Use of calculators is not permitted. However, you may use log tables if necessary.

Question 1

Question 1

Questions

Q 1Q 1

Write the expression�

for the de Broglie wavelength associated with a charged particle�

having charge ‘q’ and mass ‘m’, when it is�

accelerated by a potential V. ( 1)( 1)

Solution:

The charged particle�

has a mass m and charge q.

The kinetic energy of�

the particle is equal to the work done on it by the electric field.

Q 2Q 2

The graph shown in the�

figure represents a plot of current versus voltage for a given�

semiconductor. Identify the region, if any, over which the�

semiconductor has a negative resistance. ( 1)( 1)

Solution:

Resistance of a�

material can be found out by the slope of the curve V vs. I. Part BC�

of the curve shows the negative resistance as with the increase in�

current, voltage decreases.

Q 3Q 3

Two charges of�

magnitudes +4Q and − Q are located at points (a, 0) and (−�

3a, 0) respectively. What is the electric flux due to these charges�

through a sphere of radius ‘2a’ with its centre at the�

origin? ( 1)( 1)

Solution:

Gauss’ theorem�

states that the electric flux through a closed surface enclosing a�

charge is equal to �

�

times the magnitude of the charge enclosed.

The sphere enclose�

charge = +4Q.

Thus,

Q 4Q 4

The motion of copper�

plate is damped when it is allowed to oscillate between the two poles�

of a magnet. What is the cause of this damping? ( 1)( 1)

Solution:

As the copper plates�

oscillate in the magnetic field between the two poles of the magnet,�

there is a continuous change of magnetic field flux linked with the�

pendulum. Due to this, eddy currents are set up in the copper plate�

which try to oppose the motion of the pendulum according to the�

Lenz’s law and finally bring it to rest.

Q 5Q 5

Two identical cells,�

each of emf E, having negligible internal resistance, are connected�

in parallel with each other across an external resistance R. What is�

the current through this resistance? ( 1)( 1)

Solution:

The cells are arranged�

as shown in the circuit diagram given below:

As the internal�

resistance is negligible, so total resistance of the circuit = R

Potential difference�

across the resistance = E

So, current through the�

resistance, .

Q 6Q 6

How does the mutual�

inductance of a pair of coils change when

(i) distance between�

the coils is decreased and

(ii) number of turns in�

the coils is decreased? ( 1)( 1)

Solution:

(i) As, f = MI,�

with the decrease in the distance between the coils the magnetic flux�

linked with the secondary coil increases and hence the mutual�

inductance of the two coils will increase.

(ii) Mutual inductance�

of two coils can be found out by, M = µon1n2Al,�

i.e. M ∝ n1n2,�

so, with the decrease in number of turns mutual inductance decreases.

Q 7Q 7

Define the activity of�

a given radioactive substance. Write its S.I. unit. ( 1)( 1)

Solution:

The total decay rate of�

a radioactive sample is called the activity of the sample. The S.I.�

unit of activity is Becquerel (Bq).

Q 8Q 8

Welders wear special�

goggles or face masks with glass windows to protect their eyes from�

electromagnetic radiations. Name the radiations and write the range�

of their frequency. ( 1)( 1)

Solution:

Welders wear special�

goggles with glass windows to protect the eyes from Ultra Violet rays�

(UV rays). The range of UV rays is 4×10-7 m (400 nm)�

to 6×10-10 m (0.6 nm).

Q 9Q 9

In the circuit shown in�

the figure, identify the equivalent gate of the circuit and make its�

truth table. ( 2)( 2)

Solution:

The equivalent gate of�

the given circuit is AND gate.

Truth table:

A B A’ B’ Y’ Y1 1 0 0 0 11 0 0 1 1 00 1 1 0 1 00 0 1 1 1 0

Q 10Q 10

Explain the term ‘drift�

velocity’ of electrons in conductor. Hence obtain the�

expression for the current through a conductor in terms of ‘drift�

velocity’. ( 2)( 2)

O RO R

Describe briefly, with�

the help of a circuit diagram, how a potentiometer is used to�

determine the internal resistance of a cell.

Solution:

‘Drift�

velocity’ of electrons in a conductor: Metals contain a�

large number of free electrons. These electrons are in continuous�

random motion. Due to the random motion, the free electrons collide�

with positive metal ions with high frequency and undergo change in�

direction at each collision. So the average velocity for the�

electrons in a conductor is zero.

Now, when this�

conductor is connected to a source of emf, an electric field is�

established in the conductor, such that E = V/L

Where V= potential�

difference across the conductor and L = length of the conductor.

The electric field�

exerts an electrostatic force ‘-Ee’ on each free electron�

in the conductor. The acceleration of each electron is given by

Where e = electric�

charge on electron and m = mass of electron

The negative sign�

indicates that the force and hence the acceleration is in a direction�

opposite to the direction of the electric field. Due to this�

acceleration, the electrons attain a velocity in addition to thermal�

velocity in the direction opposite to that of electric field.

The average velocity of�

all the free electrons in the conductor is called the drift velocity�

of free electrons of the conductor.

Thus, the expression�

for the drift velocity is

where �

=�

relaxation time between two successive collision.

Let n = number density�

of electrons in the conductor.

No. of free electrons�

in the conductor = nAL

Total charge on the�

conductor, q = nALe

Time taken by this�

charge to cover the length L of the conductor, �

O RO R

Measurement�

of internal resistance of a cell using potentiometer:

The�

cell of emf, E (internal�

resistance r)�

is connected across a resistance box (R)�

through key K2.

When�

K2�

is open, balance�

length is obtained at length AN1 = l1

E= F l1……………………………………………………………………………..�

(1)

When�

K2�

is closed:

Let V be�

the terminal potential difference of cell and the balance is obtained�

at AN2 = l2

\ V�

= F l2�

……………………………………………………………………………(2)

From�

equations (1) and (2), we get

From�

(3) and (4), we get

We know l1,�

l2 and R, so we can calculate r.

Q 11Q 11

A parallel beam of�

light of 450 nm falls on a narrow slit and the resulting diffraction�

pattern is observed on a screen 1.5 m away. It is observed that the�

first minimum is at a distance of 3 mm from the centre of the screen.�

Calculate the width of the slit. ( 2)( 2)

Solution:

The distance of the nth�

minimum from the center of the screen is, �

Where, D = distance of�

slit from screen, l = wavelength of the light, a = width of�

the slit

For first minimum, n =�

1:

Q 12Q 12

In the block diagram of�

a simple modulator for obtaining an AM signal, shown in the figure,�

identify the boxes A and B. Write their function. ( 2)( 2)

Solution:

In the block diagram of�

modulator, A is Square Law Device and B is Band pass filter.

Band pass filte rBand pass filte r�

rejects low and high frequencies and allows a band of frequencies to�

pass through.

Square law de viceSquare law de vice �

is a non linear device. It produces a non-linear output of message�

and carrier signals. The output from square law device is,�

y(t)=Bx(t)+Cx2(t)

Where, B and C are�

constants and x(t)=message signal (Amsinwmt)�

+ carrier signal (Acsinwct)

Q 13Q 13

A convex lens of focal�

length f1 is kept in contact with a concave lens of�

focal length f2. Find the focal length of the�

combination. ( 2)( 2)

Solution:

For convex lens, focal�

length, f = f1 and for concave lens, the�

focal length, f = −f2

The equivalent focal�

length of a combination of convex lens and concave lens is given as:

Q 14Q 14

Draw typical output�

characteristics of an n-p-n transistor in CE configuration. Show how�

these characteristics can be used to determine output resistance. ( 2)( 2)

Solution:

Output characteristics�

is the plot between collector-emitter voltage (VCE) and�

the collector current (IC) at different constant values of�

base current (IB).

Output resistance is�

defined as the ratio of the change in collector-emitter voltage�

(ΔVCE) to the change in collector current (ΔIC)�

at a constant base current (IB).

Initially with the�

increase in VCE the collector current increases almost linearly this�

is because the junction is not reverse biased. When the supply is�

more than required to reverse bias the base-collector junction, IC�

increases very little with VCE.

The reciprocal of slope�

of the linear part of the curve gives the value of output resistance�

i.e. �

Q 15Q 15

A slab of material of�

dielectric constant K has the same area as that of the plates of a�

parallel plate capacitor but has the thickness 2d/3, where d is the�

separation between the plates. Find out the expression for its�

capacitance when the slab is inserted between the plates of the�

capacitor. ( 2)( 2)

Solution:

Initially when there is�

vacuum between the two plates, the capacitance of the two parallel�

plates is, �

,�

where, A is the area of parallel plates.

Suppose that the�

capacitor is connected to a battery, an electric field E0�

is produced.

Now if we insert the�

dielectric slab of thickness �

the�

electric field reduces to E.

Now the gap between�

plates is divided in two parts, for distance t there is electric�

field E and for the remaining distance (d–t) the electric field�

is E0 .

If V be the potential�

difference between the plates of the capacitor, then V=Et + E0(d–t)

Q 16Q 16

A capacitor, made of�

two parallel plates each of plate area A and separation d, is being�

charged by an external ac source. Show that the displacement current�

inside the capacitor is the same as the current charging the�

capacitor. ( 2)( 2)

Solution:

Let the alternating emf�

charging the plates of capacitor be E = Eo sin wt

Charge on the�

capacitor, q = EC = CEo sin wt

Instantaneous Current�

is I.

Thus, the displacement�

current inside the capacitor is the same as the current charging the�

capacitor.

Q 17Q 17

A wire AB is carrying a�

steady current of 6 A and is lying on the table. Another wire CD�

carrying 4 A is held directly above AB at a height of 1 mm. Find the�

mass per unit length of the wire CD so that it remains suspended at�

its position when left free. Give the direction of the current�

flowing in CD with respect to that in AB. [Take the value of g = 10�

ms−2] ( 3)( 3)

Solution:

Force per unit length�

between the current carrying wires is given as:

Let m be the mass per�

unit length of wire CD. As the force balances the weight of the wire

Q 18Q 18

(a) For a given a.c., i�

= im sin wt, show�

that the average power dissipated in a resistor R over a complete�

cycle is �

�

R.

(b) A light bulb is�

rated at 125 W for a 250 V a.c. supply. Calculate the resistance of�

the bulb. ( 3)( 3)

Solution:

(a) �

(b)

Power of the bulb, P =�

125 W, Voltage, V = 250 V

Q 19Q 19

Draw V − I�

characteristics of a p-n junction diode. Answer the following�

questions, giving reasons:

(i) Why is the current�

under reverse bias almost independent of the applied potential up to�

a critical voltage?

(ii) Why does the�

reverse current show a sudden increase at the critical voltage?

Name any semiconductor�

device which operates under the reverse bias in the breakdown region.�

( 3)( 3)

Solution:

V-I characteristic of�

p-n junction diode:

(i) Under the reverse�

bias condition, the holes of p-side are attracted towards the�

negative terminal of the battery and the electrons of the n-side are�

attracted towards the positive terminal of the battery. This�

increases the depletion layer and the potential barrier. However the�

minority charge carriers are drifted across the junction producing a�

small current. At any temperature the number of minority carriers is�

constant so there is the small current at any applied potential. This�

is the reason for the current under reverse bias to be almost�

independent of applied potential. At the critical voltage, avalanche�

breakdown takes place which results in a sudden flow of large�

current.

(ii) At the critical�

voltage, the holes in the n-side and conduction electrons in the�

p-side are accelerated due to the reverse −bias voltage. These�

minority carriers acquire sufficient kinetic energy from the electric�

field and collide with a valence electron. Thus the bond is finally�

broken and the valence electrons move into the conduction band�

resulting in enormous flow of electrons and thus formation of�

hole-electron pairs. Thus there is a sudden increase in the current�

at the critical voltage.

Zener diode is a�

semiconductor device which operates under the reverse bias in the�

breakdown region.

Q 20Q 20

Define the current�

sensitivity of a galvanometer. Write its S.I. unit.

Figure shows two�

circuits each having a galvanometer and a battery of 3V.

When the galvanometers�

in each arrangement do not show any deflection, obtain the ratio�

R1/R2. ( 3)( 3)

Solution:

Current sensitivity of�

a galvanometer is defined as the deflection in galvanometer per unit�

current. Its SI unit is radians/Ampere.

For balanced wheat�

stone bridge, there will be no deflection in the galvanometer.

For the equivalent�

circuit, when the wheat stone bridge is balanced, there will be no�

deflection in the galvanometer.

Q 21Q 21

A rectangular conductor�

LMNO is placed in a uniform magnetic field of 0.5 T. The field is�

directed perpendicular to the plane of the conductor. When the arm MN�

of length of 20 cm is moved towards left with a velocity of 10 ms−1,�

calculate the emf induced in the arm. Given the resistance of the arm�

to be 5 W (assuming that�

other arms are of negligible resistance) find the value of the�

current in the arm. ( 3)( 3)

O RO R

A wheel with 8 metallic�

spokes each 50 cm long is rotated with a speed of 120 rev/min in a�

plane normal to the horizontal component of the Earth’s�

magnetic field. The Earth’s magnetic field at the place is 0.4�

G and the angle of dip is 60°. Calculate the emf induced between�

the axle and the rim of the wheel. How will the value of emf be�

affected if the number of spokes were increased?

Solution:

Let ON be x at some�

instant.

The emf induced in the�

loop = e.

O RO R

This component is�

parallel to the plane of the wheel. Thus, the emf induced is given�

as,

The value of emf�

induced is independent of the number of spokes as the emf’s�

across the spokes are in parallel. So, the emf will be unaffected�

with the increase in spokes.

Q 22Q 22

(a) What is linearly�

polarized light? Describe briefly using a diagram how sunlight is�

polarised.

(b) Unpolarised light�

is incident on a polaroid. How would the intensity of transmitted�

light change when the polaroid is rotated? ( 3)( 3)

Solution:

(a) Natural light is�

unpolarised i.e. the electric vector takes all possible directions in�

the transverse plane, rapidly and randomly, during a measurement. A�

polarizer transmits only one component (parallel to a special axis).�

This resulting light is called linear or plane polarized.

The incident sunlight�

is unpolarised. The dot and double arrows show the polarization in�

the perpendicular and in the plane of the figure. Under the influence�

of the electric field of the incident wave, the electrons in the�

molecules of the atmosphere acquire components of motion in both�

these directions. An observer looking at 90° to the direction of�

the sun, the charges accelerating parallel to the double arrows do�

not radiate energy towards this observer since their acceleration has�

no transverse component. The radiation scattered by the molecule is�

therefore represented by dots. It is linearly polarized perpendicular�

to the plane of the figure.

(b) If the unpolarised�

light is incident on a Polaroid the intensity is reduced by half.�

Even if the Polaroid is rotated by angle q the average over�

cos2q = ½. Thus from Malus’ law: I =�

I0cos2q

Or, <I> = <�

I0cos2q > = I0< cos2q�

> = I0/2

Thus, the intensity of�

the transmitted light remains unchanged when the Polaroid is rotated.

Q 23Q 23

Draw a labelled ray�

diagram of a refracting telescope. Define its magnifying power and�

write the expression for it.

Write two important�

limitations of a refracting telescope over a reflecting type�

telescope. ( 3)( 3)

Solution:

Refracting�

telescope:

Magnifying P o we r:Magnifying P o we r:�

The magnifying power m is the ratio of the angle�

a subtended at the eye�

by the final image to the angle b which the object subtends at�

the lens or the eye.

Limitatio ns o f�Limitatio ns o f�

re fracting te le sco pe o ve r re fle cting type te le sco pe :re fracting te le sco pe o ve r re fle cting type te le sco pe :

(NOTE: Write any two)

(i) Refracting�

telescope suffers from chromatic aberration as it uses large sized�

lenses.

(ii) The image formed�

by refracting telescope is less brighter than the image formed by the�

reflecting type telescope due to some loss of light by reflection at�

the lens and by absorption.

(iii) The resolving�

power of refracting telescope is less than the resolving power of�

reflecting type telescope as the mirror of reflecting type telescope�

has large diameter.

(iv) The requirements�

of big lenses tend to be very heavy and therefore difficult to make�

and support by their edges.

(v) It is also�

difficult and expensive to make such large sized lenses.

Q 24Q 24

Write Einstein’s�

photoelectric equation and point out any two characteristic�

properties of photons on which this equation is based.

Briefly explain the�

three observed features which can be explained by this equation. ( 3)( 3)

Solution:

Einstein’s�

photoelectric effect equation:

( NO TE: Use any o ne �( NO TE: Use any o ne �

o f thre e give n e quatio ns)o f thre e give n e quatio ns)

O RO R

O RO R

The two�

characteristic properties of photons on which this equation is based�

are as follows:

(i) Photons have�

particle characteristic. It is emitted or absorbed in units called�

quanta of light.

(ii) Photons have wave�

characteristic. It travels in space with particular frequency, a�

characteristic of waves.

Three observed�

features which can be explained by this equation are:

(i) Solar cells: Also�

called photo-voltaic cells. It converts solar radiations to�

electrical emf.

(ii) Television�

telecast: The dark and bright light part of images are interpreted as�

high and low electrical charges as given by photoelectric emission�

principle. These are further processed and transmitted.

(iii) Burglar alarm:�

The moment the ultraviolet radiation is cut due to theif, it stops�

the supply of photons and thus works as ‘off’ mode and�

the ring bells automatically.

Q 25Q 25

Name the type of waves�

which are used for line of sight (LOS) communication. What is the�

range of their frequencies?

A transmitting antenna�

at the tope of a tower has a height of 45 m and the receiving antenna�

is on the ground. Calculate the maximum distance between them for�

satisfactory communication in LOS mode. (Radius of the Earth = 6.4 �

106 m) ( 3)( 3)

Solution:

Space waves are used�

for the line of sight (LOS) communication.

The range of their�

frequencies is 40 MHz and above.

We have, height of�

transmitting antenna, hT = 45 m and height of receiving�

antenna, hR = 0 m

Then, maximum distance�

between the two antennas, �

Thus, the maximum�

distance between the antennas is 24 km.

Q 26Q 26

One day Chetan’s�

mother developed a severe stomach ache all of a sudden. She was�

rushed to the doctor who suggested for an immediate endoscopy test�

and gave an estimate of expenditure for the same. Chetan immediately�

contacted his class teacher and shared the information with her. The�

class teacher arranged for the money and rushed to the hospital. On�

realizing that Chetan belonged to a below average income group�

family, even the doctor offered concession for the test fee. The test�

was conducted successfully.

Answer the following�

questions based on the above information:

(a) Which principle in�

optics is made use of in endoscopy?

(b) Briefly explain the�

values reflected in the action taken by the teacher.

(c) In what way do you�

appreciate the response of the doctor on the given situation? ( 4)( 4)

Solution:

(a) Endoscopy is based�

on total internal reflection principle. It has tubes which are made�

up of optical fibres which are used for transmitting and receiving�

electrical signals which are converted to light by suitable�

transducer.

(b) Humanity and�

Charity.

(c) Doctor gave�

monetary help to Chetan by understanding his poor financial�

condition.

Q 27Q 27

(a) Define electric�

dipole moment. Is it a scalar or a vector? Derive the expression for�

the electric field of a dipole at a point on the equatorial plane of�

the dipole.

(b) Draw the�

equipotential surfaces due to an electric dipole. Locate the points�

where the potential due to the dipole is zero.

O RO R

Using Gauss’ law�

deduce the expression for the electric field due to a uniformly�

charged spherical conducting shell of radius R at a point

(i) outside and (ii)�

inside the shell.

Plot a graph showing�

variation of electric field as a function of r > R and r < R.

(r being the distance�

from the centre of the shell) ( 5)( 5)

Solution:

(a) Electric dipole�

moment: The strength of an electric dipole is measured by the�

quantity electric dipole moment. Its magnitude is equal to the�

product of the magnitude of either charge and the distance between�

the two charges.

Electric dipole moment,�

p = q × d

It is a vector�

quantity.

In vector form it is�

written as ,�

where the direction of �

is�

from negative charge to positive charge.

Electric Field�

of dipole at points on the equatorial plane:

The magnitudes of�

the electric field due to the two charges +q and −q are�

given by,

The directions�

of E+q and E−q are�

as shown in the figure. The components normal to the dipole axis�

cancel away. The components along the dipole axis add up.

\ Total�

electric field

[Negative�

sign shows that field is opposite to ]

At large distances�

(r >> a), this reduces to

(b) Equipotential�

surface due to electric dipole:

The potential due to�

the dipole is zero at the line bisecting the dipole length.

O RO R

Electric Field�

Due To A Uniformly Charged Thin Spherical Shell:

(i) When point�

P lies outside the spherical shell:

Suppose that we�

have to calculate electric field at the point P at a�

distance r (r > R)�

from its centre. Draw the Gaussian surface through point P so as to�

enclose the charged spherical shell. The Gaussian surface is a�

spherical shell of radius r and centre O.

Let be�

the electric field at point P. Then, the electric flux through area�

element is�

given by,

Since is�

also along normal to the surface,

dF = E ds

\ Total�

electric flux through the Gaussian surface is given by,

Now,

Since the charge�

enclosed by the Gaussian surface is q, according to Gauss�

theorem,

From equations (i)�

and (ii), we obtain

(ii) When point�

P lies inside the spherical shell:

In such a case,�

the Gaussian surface encloses no charge.

According to Gauss�

law,

E × 4pr2 =�

0

i.e., = E =�

0 (r < R)

Graph showing the�

variation of electric field as a function of r:

Q 28Q 28

Using Bohr’s�

postulates, derive the expression for the frequency of radiation�

emitted when electron in hydrogen atom undergoes transition from�

higher energy state (quantum number ni) to the lower�

state, (nf).

When electron in�

hydrogen atom jumps from energy state ni = 4 to nf�

= 3, 2, 1, identify the spectral series to which the emission lines�

belong. ( 5)( 5)

O RO R

(a) Draw the plot of�

binding energy per nucleon (BE/A) as a functino of mass number A.�

Write two important conclusions that can be drawn regarding the�

nature of nuclear force.

(b) Use this graph to�

explain the release of energy in both the processes of nuclear fusion�

and fission.

(c) Write the basic�

nuclear process of neutron undergoing b-decay.

Why is the detection of�

neutrinos found very difficult?

Solution:

In the hydrogen atom,

Now,�

according to Bohr’s frequency condition when electron in�

hydrogen atom undergoes transition from higher energy state (quantum�

number ni) to the lower state (nf) is,

Now,higher�

state ni = 4, lower state, nf = 3, 2, 1

For�

the transition,

ni�

= 4 to nf = 3:® Paschen series

ni�

= 4 to nf = 2:® Balmer series

ni�

= 4 to nf = 1:® Lyman series

O RO R

(a) Plot of binding�

energy per nucleon as the function of mass number A is given as�

below:

Following are the two�

conclusions that can be drawn regarding the nature of the nuclear�

force:

(i)�The force is attractive and strong enough to produce a binding energy�of few MeV per nucleon.(ii)�The constancy of the binding energy in the range 30<A<170 is a�consequence of the fact that the nuclear force is short range force.( b) ( b) Nucle ar�Nucle ar�

fissio n:fissio n: A very heavy nucleus (say A = 240) has lower binding�

energy per nucleon as compared to the nucleus with A = 120. Thus if�

the heavier nucleus breaks to the lighter nucleus with high binding�

energy per nucleon, nucleons are tightly bound. This implies that�

energy will be released in the process which justifies the energy�

release in fission reaction.

Nucle ar fusio n: Nucle ar fusio n: When�

two light nuclei (A<10) are combined to form a heavier nuclei, the�

binding energy of the fused heavier nuclei is more than the binding�

energy per nucleon of the lighter nuclei. Thus the final system is�

more tightly bound than the initial system. Again the energy will be�

released in fusion reaction.

( c)( c) The basic�

nuclear process of neutron undergoing b-decay is given as:

Neutrinos interact very�

weakly with matter so, they have a very high penetrating power.�

That’s why the detection of neutrinos is found very difficult.

Q 29Q 29

(a) Using Biot −�

Savart’s law, derive the expression for the magnetic field in�

the vector form at a point on the axis of a circular current loop.

(b) What does a toroid�

consist of? Find out the expression for the magnetic field inside a�

toroid for N turns of the coil having the average radius r and�

carrying a current I. Show that the magnetic field in the open space�

inside and exterior to the toroid is zero.

O RO R

(a) Draw a schematic�

sketch of a cyclotron. Explain clearly the role of crossed electric�

and magnetic field in accelerating the charge. Hence derive the�

expression for the kinetic energy acquired by the particles.

(b) An a-particle�

and a proton are released from the centre of the cyclotron and made�

to accelerate.

(i) Can both be�

accelerated at the same cyclotron frequency?

Give reason to justify�

your answer.

(ii) When they are�

accelerated in turn, which of the two will have higher velocity at�

the exit slit of the does? ( 5)( 5)

Solution:

( a)( a)�

Magne tic fie ld�Magne tic fie ld�

o n the axis o f a circular curre nt lo o po n the axis o f a circular curre nt lo o p

I ®Current�

in the loop, R ®Radii�

of the loop, X-axis�

®Axis of the loop, x ®Distance�

of OP

dl ®Conducting�

element of the loop

According�

to Biot-Savart’s law, the magnetic field at P is �

r2�

= x2+ R2

| dl × r |�

= r dl ( they�

are perpendicular)

dB has�

two components − dBx and dB^. dB^ is�

cancelled out and only the x-component remains.

\ dBx= dBcosq

Cosq�

=

\ dBx=

Summation�

of dl over the loop is given by 2pR.

\ B =

( b)( b)�

Toroid is a hollow circular ring on�

which a large number of turns of a wire are closely wound.

Figure shows a�

sectional view of the toroid. The direction of the magnetic field�

inside is clockwise as per the right-hand thumb rule for circular�

loops. Three circular Amperian loops 1, 2 and 3 are shown by dashed�

lines.

By symmetry, the�

magnetic field should be tangential to each of them and constant in�

magnitude for a given loop.

Let the magnetic field�

inside the toroid be B. We shall now consider the magnetic�

field at S. Once again we employ

Ampere’s law in�

the form of,

Or,�

BL = µ0NI

Where, L is the length�

of the loop for which B is tangential, I be the current enclosed by�

the loop and N be the number of turns.

We find, L = 2pr.

The current enclosed I�

is (for N turns of toroidal coil) N I.

B (2pr) = µ0NI�

, therefore, �

Open space inside�

the toroid encloses no current thus, I = 0.

Hence, B = 0

Open space�

exterior to the toroid:

Each turn of current�

carrying wire is cut twice by the loop 3. Thus, the current coming�

out of the plane of the paper is cancelled exactly by the current�

going into it. Thus,

I= 0, and B = 0.

O RO R

( a)( a) Schematic�

sketch of cyclotron:

The combination of�

crossed electric and magnetic fields is used to increase the energy�

of the charged particle. Cyclotron uses the fact that the frequency�

of revolution of the charged particle in a magnetic field is�

independent of its energy. Inside the dees the particle is shielded�

from the electric field and magnetic field acts on the particle and�

makes it to go round in a circular path inside a dee. Every time the�

particle moves from one dee to the other it comes under the influence�

of electric field which ensures to increase the energy of the�

particle as the sign of the electric field changed alternately. The�

increased energy increases the radius of the circular path so the�

accelerated particle moves in a spiral path.

Since�

radius of trajectory,

Hence,�

the kinetic energy of ions,

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( b)( b)

(i) Let us consider:�

Mass of proton = m, Charge of proton = q, Mass of alpha particle = 4m

Charge of alpha�

particle = 2q

Thus, particles will�

not accelerate with same cyclotron frequency. The frequency of proton�

is twice than the frequency of alpha particle.

(ii)

Thus particles will not�

exit the dees with same velocity. The velocity of proton is twice�

than the velocity of alpha particle.

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